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Thread: Euler Identity Algebra Problem

  1. #1
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    Euler Identity Algebra Problem

    Using Euler's identity, find real positive constants $\displaystyle c$ and $\displaystyle \phi$ for all real $\displaystyle t$ such that:

    $\displaystyle 3 \cos (2t) - 4 \sin (2t + \frac{\pi}{4}) = c \cos (2t + \phi)$

    I tried substituting the Euler formula for all sine and cosine values to get:

    $\displaystyle \frac{3}{2}(e^{j2t}+e^{-j2t}) - \frac{4}{2}(e^{j2t}e^{j\pi /4}-e^{-j2t}e^{-j\pi /4}) = \frac{c}{2}(e^{j2t}e^{j\phi}+e^{-j2t}e^{-j\phi})$

    Which suggests:

    $\displaystyle ce^{j\phi} = 3 - 4 e^{j\pi /4}$

    Which I don't see a solution to. Any ideas?
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  2. #2
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    Re: Euler Identity Algebra Problem

    Quote Originally Posted by VinceW View Post
    Using Euler's identity, find real positive constants $\displaystyle c$ and $\displaystyle \phi$ for all real $\displaystyle t$ such that:

    $\displaystyle 3 \cos (2t) - 4 \sin (2t + \frac{\pi}{4}) = c \cos (2t + \phi)$

    I tried substituting the Euler formula for all sine and cosine values to get:

    $\displaystyle \frac{3}{2}(e^{j2t}+e^{-j2t}) - \frac{4}{2}(e^{j2t}e^{j\pi /4}-e^{-j2t}e^{-j\pi /4}) = \frac{c}{2}(e^{j2t}e^{j\phi}+e^{-j2t}e^{-j\phi})$

    Which suggests:

    $\displaystyle ce^{j\phi} = 3 - 4 e^{j\pi /4}$

    Which I don't see a solution to. Any ideas?
    $\displaystyle \displaystyle \begin{align*} 3\cos{(2t)} - 4\sin{\left(2t + \frac{\pi}{4}\right)} &= 3\cos{(2t)} - 4\left[\sin{(2t)}\cos{\left(\frac{\pi}{4}\right)} + \cos{(2t)}\sin{\left(\frac{\pi}{4}\right)}\right] \\ &= 3\cos{(2t)} - 4\left[\frac{\sqrt{2}}{2}\sin{(2t)} + \frac{\sqrt{2}}{2}\cos{(2t)}\right] \\ &= 3\cos{(2t)} - 2\sqrt{2}\sin{(2t)} - 2\sqrt{2}\cos{(2t)} \\ &= \left(3 - 2\sqrt{2}\right)\cos{(2t)} - 2\sqrt{2}\sin{(2t)} \end{align*}$

    And...

    $\displaystyle \displaystyle \begin{align*} c\cos{\left(2t + \phi\right)} &= c\left[\cos{(2t)}\cos{(\phi)} - \sin{(2t)}\sin{(\phi)}\right] \\ &= c\cos{(\phi)}\cos{(2t)} - c\sin{(\phi)}\sin{(2t)} \end{align*}$

    So

    $\displaystyle \displaystyle c\cos{(\phi)} = 3 - 2\sqrt{2} \\ \implies c^2\cos^2{(\phi)} = \left(3 - 2\sqrt{2}\right)^2 = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2}$

    and $\displaystyle \displaystyle c\sin{(\phi)} = 2\sqrt{2} \implies c^2\sin^2{(\phi)} = 8$.

    Therefore

    $\displaystyle \displaystyle \begin{align*} c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 2\sqrt{2} \\ c^2\left[\cos^2{(\phi)} + \sin^2{(\phi)} \right] &= 17 - 10\sqrt{2} \\ c^2(1) &= 17 - 10\sqrt{2} \\ c^2 &= 17 - 10\sqrt{2} \\ c &= \pm \sqrt{17 - 10\sqrt{2}} \end{align*}$

    Can you go from here to get values for $\displaystyle \displaystyle \phi$?
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  3. #3
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    Re: Euler Identity Algebra Problem

    Thanks Prove It. You definitely solved it. You didn't use the Euler formula, but I guess that wasn't applicable?

    I believe you made a small algebra error in that this:

    $\displaystyle c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 2\sqrt{2}$

    should be:

    $\displaystyle c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 8$

    thanks
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    Re: Euler Identity Algebra Problem

    Quote Originally Posted by VinceW View Post
    Thanks Prove It. You definitely solved it. You didn't use the Euler formula, but I guess that wasn't applicable?

    I believe you made a small algebra error in that this:

    $\displaystyle c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 2\sqrt{2}$

    should be:

    $\displaystyle c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 8$

    thanks
    Yes you are correct, that was a typo.

    Did you manage to find any values for $\displaystyle \displaystyle \phi$?
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  5. #5
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    Re: Euler Identity Algebra Problem

    Quote Originally Posted by Prove It View Post
    Did you manage to find any values for $\displaystyle \displaystyle \phi$?
    Yes, that was last step was obvious after all the previous steps:

    $\displaystyle \phi = \arccos \frac{3 - 2\sqrt{2}}{\sqrt{25 - 12\sqrt{2}}} = \arcsin \frac{2\sqrt{2}}{\sqrt{25 - 12\sqrt{2}}} \approx 1.510$
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